(B) $\left(\dfrac{11}{7}, 0\right)$
Using section formula: $x = \dfrac{5(1)+2(4)}{5+2} = \dfrac{13}{7}$... wait — $x = \dfrac{5(1)+2(4)}{7} = \dfrac{13}{7}$, $y = \dfrac{5(2)+2(-5)}{7} = \dfrac{0}{7} = 0$. So P = $\left(\dfrac{13}{7}, 0\right)$. (C)
Apply the section formula with $m_1:m_2 = 5:2$, $A(4,-5)$, $B(1,2)$:
$$x = \frac{5(1)+2(4)}{7} = \frac{13}{7}, \quad y = \frac{5(2)+2(-5)}{7} = \frac{0}{7} = 0$$
The correct answer is (C) $\left(\dfrac{13}{7}, 0\right)$. Option (B) has a wrong x-value ($\frac{11}{7}$); don't confuse $5×1+2×4 = 13$ with any other combination. Always substitute carefully: $m_1$ goes with $x_2$ (coordinates of B), and $m_2$ goes with $x_1$ (coordinates of A).